Questions: [See below]
$\rm\color{#c00}{(1)}$ What structure has $R/I$ when $I$ is a maximal left ideal? I know that if $I$ is a bilateral ideal then $R/I$ is a ring and, moreover, if $I$ is maximal then it is a field, in which case we get the simplicity of $R/I$.
This question naturally gives rise to the second question.
$\rm\color{#c00}{(2)}$ I know that if we define $J'(R)$ by changing the word left for the word right in $(\star)$, then $J(R)=J'(R)$ and the characterization $(\star\star)$ remains valid, with the word left changed for the word right. However, can we put the word bilateral instead of left or right in $(\star\star)$? In the French Wikipedia page Radical de Jacobson, $J(R)$ seems to be defined exactly this way and the proof of $(\star\star\star)$ presented there seems to make more sense since in that case $R/I$ is a field as usual.
Define the Jacobson radical of a ring (with unity) $R$ by $$ J(R):=\bigcap_{M\text{ a simple, left }R\text{-module}}\text{Ann}_R(M).\quad(\star) $$ Then one can show $$ J(R)=\bigcap_{I\text{ a maximal, left ideal of }R}I.\quad(\star\star) $$ Next, in order to show the $[\Longleftarrow]$ part of $$ r\in J(R)\iff\forall x\in R:1+xr\in R^{\times},\quad(\star\star\star) $$ one proceeds by contradiction as follows:
Proof of $[\Longleftarrow]$: Suppose that for all $x\in R$ we have $1+xr\in R^{\times}$ but that there is a maximal left ideal $I$ of $R$ for which $r\not\in I$. $\rm\color{#c00}{(1)}$ Then $R/I$ is simple, hence $R\overline{r}=I\text{ or }R/I$. Since $r\not\in I$, $R\overline{r}=R/I$ and so there is an $x\in R$ such that $1+xr\in I$. But $1+xr$ is invertible, hence $1\in I$ and $I=R$ which contradicts the hypothesis that $I$ is maximal. $\blacksquare$
The best we can say is that it is a left $R$ module and not necessarily anything more.
No, in general you will only get a simple ring.
It will not yield the Jacobson radical. Let $R$ be the ring of linear transformations of a countable dimensional vector space. Then $R$ has exactly one nontrivial bilateral ideal: the set of transformations with finite dimensional range. However the whole ring is Von Neumann regular with Jacobson radical zero.
I notice that the article says "le radical de Jacobson d'un anneau commutatif est..." Of course if the ring is commutative then we don't need left or right ideals anymore.